\(\int \sec ^4(c+d x) (a+b \sin (c+d x)) \, dx\) [385]

Optimal result

Integrand size = 19, antiderivative size = 44 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {b \sec ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \]

[Out]

1/3*b*sec(d*x+c)^3/d+a*tan(d*x+c)/d+1/3*a*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2748, 3852} \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {b \sec ^3(c+d x)}{3 d} \]

[In]

Int[Sec[c + d*x]^4*(a + b*Sin[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^3)/(3*d) + (a*Tan[c + d*x])/d + (a*Tan[c + d*x]^3)/(3*d)

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b \sec ^3(c+d x)}{3 d}+a \int \sec ^4(c+d x) \, dx \\ & = \frac {b \sec ^3(c+d x)}{3 d}-\frac {a \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d} \\ & = \frac {b \sec ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {b \sec ^3(c+d x)}{3 d}+\frac {a \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \]

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sin[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^3)/(3*d) + (a*(Tan[c + d*x] + Tan[c + d*x]^3/3))/d

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {b}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(38\)
default	\(\frac {-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {b}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(38\)
risch	\(\frac {4 i {\mathrm e}^{2 i \left (d x +c \right )} a +\frac {8 b \,{\mathrm e}^{3 i \left (d x +c \right )}}{3}+\frac {4 i a}{3}}{d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{3}}\)	\(49\)
parallelrisch	\(\frac {-6 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +4 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\)	\(90\)
norman	\(\frac {-\frac {2 b}{3 d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(156\)

[In]

int(sec(d*x+c)^4*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-a*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+1/3*b/cos(d*x+c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (2 \, a \cos \left (d x + c\right )^{2} + a\right )} \sin \left (d x + c\right ) + b}{3 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/3*((2*a*cos(d*x + c)^2 + a)*sin(d*x + c) + b)/(d*cos(d*x + c)^3)

Sympy [F]

\[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**4*(a+b*sin(d*x+c)),x)

[Out]

Integral((a + b*sin(c + d*x))*sec(c + d*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a + \frac {b}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/3*((tan(d*x + c)^3 + 3*tan(d*x + c))*a + b/cos(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.73 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {2 \, {\left (3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} d} \]

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-2/3*(3*a*tan(1/2*d*x + 1/2*c)^5 + 3*b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^3 + 3*a*tan(1/2*d*x +
1/2*c) + b)/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*d)

Mupad [B] (verification not implemented)

Time = 4.69 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\frac {2\,a\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^2}{3}+\frac {b}{3}+\frac {a\,\sin \left (c+d\,x\right )}{3}}{d\,{\cos \left (c+d\,x\right )}^3} \]

[In]

int((a + b*sin(c + d*x))/cos(c + d*x)^4,x)

[Out]

(b/3 + (a*sin(c + d*x))/3 + (2*a*cos(c + d*x)^2*sin(c + d*x))/3)/(d*cos(c + d*x)^3)